INVERSE OF A MATRIX IN MACHINE LEARNING

INVERSE OF A MATRIX. Definition. (Inverse of a Matrix). Let A be a square matrix of order n. A matrix B of order n is said to be the inverse of A if

AB=BA=I_n

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Note that the inverse of an arbitrary square matrix may not exist. For example the matrix

A=\left[\begin{array}{rr}1 & -1 \\\ -1 & 1\end{array}\right]

has no inverse. An inverse, if it exits, is unique. For, suppose that A has two inverses, say B and C. Then [katex]AB=BA=I \quad[/katex] and [katex]\quad A C=C A=I[/katex]. Therefore, by the associative law for multiplication, we have

B(A C)=B I=B

and

\quad(B A) C=I C=C

and so

\quad B=C

The unique inverse of a matrix A, if it exists, is denoted by [katex]A^{-1}[/katex]. A square matrix A, whose inverse exists, is called a nonsingular (or an invertible) matrix. Square matrices which do not have inverses are called singular matrices.

Properties

It is easy to verify the following properties of inverses of matrices: (i) For any invertible matrix [katex]A,\left(A^{-1}\right)^{-1}=A[/katex]. (ii) For nonsingular matrices A and B, [katex](A B)^{-1}=B^{-1} A^{-1}[/katex], (iii) For any invertible matrix [katex]A,\left(A^T\right)^{-1}=\left(A^{-1}\right)^T[/katex]. Proof. (ii) Since A and B are nonsingular, [katex]A^{-1}[/katex] and [katex]B^{-1}[/katex] exist. Also, since A and B are square matrices AB and [katex]B^{-1} A^{-1}[/katex] are defined. To prove that A B is nonsingular and [katex]B^{-1} A^{-1}[/katex] is the inverse of AB, we show that

(AB)\left(B^{-1} A^{-1}\right)=I=\left(B^{-1} A^{-1}\right)(AB)

Now

(A B)\left(B^{-1} A^{-1}\right)=A\left(B B^{-1}\right) A^{-1}=A IA^{-1}=A A^{-1}=I

and

\left(B^{-1} A^{-1}\right)(A B)=B^{-1}\left(A^{-1} A\right) B=B^{-1} I B=B^{-1} B=I

Thus the product AB of two non-singular matrices A and B is non singular and [katex](AB)^{-1}=B^{-1}A^{-1}[/katex].

We can generalize the above result for the product of a finite number of nonsingular matrices of the same order. Thus, if [katex]A_1, A_2, \cdots, A_m[/katex] are all nonsingular matrices of the same order, then

\left(A_1 A_2 \cdots A_m\right)^{-1}=A_m^{-1} \cdots A_2^{-1} A_1^{-1} .

So the product of any finite number of nonsingular matrices is nonsingular. Proofs of (i) and (iii) are left as an exercise. Example. Let

A=\left[\begin{array}{rrr}
6 & -2 & -3 \\
-1 & 1 & 0 \\
-1 & 1 & 1
\end{array}\right]\\

and

B=\left[\begin{array}{ccc}
1 / 4 & -1 / 4 & 3 / 4 \\
1 / 4 & 3 / 4 & 3 / 4 \\
0 & -1 & 1
\end{array}\right]

Now

AB=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I_3 \quad

and

BA=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I_3

Hence [katex]B=A^{-1}[/katex]. Also [katex]B^{-1}=A[/katex].